- #1
HRubss
- 66
- 1
- Homework Statement
- The motorcycle is traveling at 40 m/s when it is at A. If the speed is then decreased at [tex]v'=-(0.05s)m/s^2[/tex], where s is in meters measured from A, determine its speed and acceleration when it reaches B. I attached a picture of the problem.
- Relevant Equations
- [tex]S = S_0 + v_0(t) + \frac{1}{2}at^2[/tex]
[tex] v^2 = (v_0)^2 + 2a(\Delta S)[/tex]
[tex]s = r\theta[/tex]
[tex]a_n = \frac{v^2}{\rho}[/tex]
[tex]a_t = v'[/tex]
Problem Statement: The motorcycle is traveling at 40 m/s when it is at A. If the speed is then decreased at [tex]v'=-(0.05s)m/s^2[/tex], where s is in meters measured from A, determine its speed and acceleration when it reaches B. I attached a picture of the problem.
Relevant Equations: [tex]S = S_0 + v_0(t) + \frac{1}{2}at^2[/tex]
[tex] v^2 = (v_0)^2 + 2a(\Delta S)[/tex]
[tex]s = r\theta[/tex]
[tex]a_n = \frac{v^2}{\rho}[/tex]
[tex]a_t = v'[/tex]
I figured since the motorcycle travels along an arc, I needed to get the arc length. [tex] s = 150m(60*\frac{\pi}{180}) = 157.08[/tex] .
Then since the tangential acceleration is constant, using the constant acceleration formula to find final velocity...
[tex] v = \sqrt{(40)^2+2(-0.05(157.08))(157.08)}[/tex] but that gave me an imaginary number since the acceleration is negative? I'm not sure if this is the correct process. Any help is appreciated!
Relevant Equations: [tex]S = S_0 + v_0(t) + \frac{1}{2}at^2[/tex]
[tex] v^2 = (v_0)^2 + 2a(\Delta S)[/tex]
[tex]s = r\theta[/tex]
[tex]a_n = \frac{v^2}{\rho}[/tex]
[tex]a_t = v'[/tex]
I figured since the motorcycle travels along an arc, I needed to get the arc length. [tex] s = 150m(60*\frac{\pi}{180}) = 157.08[/tex] .
Then since the tangential acceleration is constant, using the constant acceleration formula to find final velocity...
[tex] v = \sqrt{(40)^2+2(-0.05(157.08))(157.08)}[/tex] but that gave me an imaginary number since the acceleration is negative? I'm not sure if this is the correct process. Any help is appreciated!